Mendelian randomization and instrumental variable regression
Dec 11, 2017 13:35 · 380 words · 2 minutes read
$$ \newcommand\cov{\text{Cov}} \newcommand\var{\text{Var}} \newcommand\iv{\text{IV}} $$
Meta data of reading
- Links:
- Year: NA
- DOI: NA
The problem and the idea of MR
Suppose we have phenotype $Y$, gene expression $X$, and genotype $Z$. The goal is to see if $X$ and $Y$ has some causal relationship. Since there are some unknown confounders, the residual of $Y \sim X$ is correlated with $X$. Therefore, the OLS estimator of effect size $\hat{\beta}_{xy}$ is biased.
To account for such drawback, $Z$ is introduced as instrumental variable (IV) since genotype is pre-determined so that there should not be confounders that can affect genotype. Therefore, the residual of $Y \sim Z$ should not be correlated with $Z$. So, OLS estimator is unbiased. The estimator constructed by $\hat{\beta}_{zy}$ and $\hat{\beta}_{zx}$ is simply:
$$\begin{aligned} \hat{\beta}_{xy} &= \frac{\hat{\beta}_{zy}}{\hat{\beta}_{zx}} \end{aligned}$$
Consistency and the derivation of the variance
The following is derived from https://www.bauer.uh.edu/rsusmel/phd/ec1-8.pdf.
Formally, IV estimator is defined as $b_{\iv} = (Z’X)^{-1} Z’y$. Then, we have:
$$\begin{aligned} b_{\iv} &= (Z’X)^{-1} Z’y \cr &= (Z’X)^{-1} Z’(X \beta_{xy} + \epsilon) \cr &= \beta_{xy} + (Z’X)^{-1} Z’\epsilon \cr &\xrightarrow{p} \beta_{xy} \end{aligned}$$
since $Z$ is independent to error.
The variance of $b_{\iv}$ is inspired by Lindeberg-Feller CLT:
$$\begin{aligned} \sqrt{N} (b_{\iv} - \beta_{xy}) &= \sqrt{N} (Z’X)^{-1} Z’\epsilon \cr &= (Z’X / N)^{-1} \sqrt{N} (Z’\epsilon / N) \cr (Z’\epsilon / N) \sqrt{N} &\xrightarrow{d} \mathcal{N}(0, \sigma^2 \var(Z)) \quad\text{, by L-F CLT} \cr \sqrt{N} (b_{\iv} - \beta_{xy}) &\xrightarrow{d} \mathcal{N}(0, \sigma^2 \cov(Z, X)^{-1} \var(Z) \cov(Z, X)^{-1}) \end{aligned}$$
The last line is kind of heuristic to me but I cannot find a justification for it (anyway …). Note that $\sigma^2 := \var(\epsilon)$ and it turns out that $\hat{\sigma}^2 = \frac{1}{N} \sum_i (y_i - b_{\iv} x_i)^2$ is an unbiased estimator of this term. So, $\hat{\var}(b_{\iv}) = \hat{\sigma}^2 \hat{\cov}(Z, X)^{-1} \hat{\var}(Z) \hat{\cov}(Z, X)^{-1}$.
The general procedure of MR
This section describes the MR procedure as stated in this paper (see post).
$b_{\iv}$ can be computed by two-step least squares (2SLS), simply $\hat{b}_{xy} = \hat{b}_{zy} / \hat{b}_{zx}$. By the result of the above section, we have:
$$\begin{aligned} \var(\hat{b}_{xy}) &= \frac{\text{unexplained variance by 2SLS}}{N} \times \frac{\var(Z)}{\cov(Z, X)^{2}} \cr &= \frac{\text{unexplained variance by 2SLS}}{N\var(X) \rho^2_{xz}} \text{, Since } \rho^2_{xz} := \frac{\cov(X, Z)^2}{\var(Z)\var(X)} \cr &= \frac{\var(Y) (1 - R_{xy}^2)}{N \var(X) R_{zx}^2} \end{aligned}$$
, which justifies the result in the paper (equation 2).